(I just need the practice problem worked out. The other picture has the background info needed to solve the answer. The correct answer is also provided to be able to check work.) Finally, weVelocity as a function of position for anaccelerationFor an object moving in a straight line with constant acceleration ax,-22v² = voz² + 2ax(x − xo).UxUnits: The units of each term in this equation reduce to m²/s².Notes:is the initial velocity of the object when it is at its initial position xo.•Vox• UxThis equation is generally used in problems that neither give a time t nor ask for oU is the final velocity of the object when it is at its final position x.EXAMPLE 2.7 Entering the freewayisNow we will apply Equation 2.11 to a problem in which time is neither given nor asked for. A sports carsitting at rest in a freeway entrance ramp. The driver sees a break in the traffic and floors the car's accelera-tor, so that the car accelerates at a constant 4.9 m/s² as it moves in a straight line onto the freeway. Whatdistance does the car travel in reaching a freeway speed of 30 m/s?SOLUTIONSET UP As shown in Figure 2.19, we place the origin at the initial posi-tion of the car and assume that the car travels in a straight line in the +xdirection. Then Vox = 0, Ux = 30 m/s, and ax 4.9 m/s².SOLVE The acceleration is constant; the problem makes no mentionof time, so we can't use Equation 2.6 or Equation 2.10 by itself.22Vv² = v₁² + 2ax(x − xo).x - xo.We need a relationship between x and Ux, and this is provided byEquation 2.11:Rearranging and substituting numerical values, we obtain(30 m/s)² - 02(4.9 m/s²)V2X-2VOX(2.112ax=Video Tutor Solut=EquatiposEcIfS92 m.CONT Vox = 0t =UxD-otxoax = 4.9 m/s²A FIGURE 2.19 A car accelerating on a ramp and merging ontoa freeway.axAlternative Solution: We could have used Equation 2.6 to solve forthe time t first:VOXx = ?30 m/s-04.9 m/s²6Ux = 30 m/s= 6.12 s.Xt accelerationeA2.4 Motion with Constant Acceleration 43Then Equation 2.10 gives the distance traveled:x = xo = Voxt + a₂²-хо= 0 + (4.9 m/s²) (6.12 s)² = 92 m.REFLECT We obtained the same result in one step when we usedEquation 2.11 and in two steps when we used Equations 2.6 and 2.10.When we use them correctly, Equations 2.6, 2.10, and 2.11 always giveconsistent results.The final speed of 30 m/s is about 67 mi/h, and 92 m is about100 yd. Does this distance correspond to your own driving experience?Practice Problem: What distance has the car traveled when it hasreached a speed of 20 m/s? Answer: 41 m.Equations 2.6, 2.10, and 2.11 are the equations of motion with constant acceleration.any kinematic problem involving the motion of an object in a straight line with constantcceleration can be solved by applying these equations. Here's a summary of the equations:The plot with constant acceleration:^ x = xo + vaxt + a 1²The effect ofacceleration:a,t²

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Newton's equations of motion can be used to solve constant acceleration problems. The acceleration…

Practice Problem: What distance has the car traveled when it has reached a speed of 20 m/s? Answer: 41 m.